Question

A straight line L with negative slope passes through the point $(8,2)$ and cuts the positive coordinates axes at points P and Q. Find the absolute minimum value of $OP+OQ$, as L varies, where O is the origin

Answer 1

Answer: 18

Let the line be
$y-2=-k(x-8)k>0$
$\text{ Point }P\left(\frac{8k+2}{k},0\right);\text{ Point }Q(0,8k+2)$
$\therefore OP+OQ=\frac{8k+2}{k}+8k+2=10+\frac{2}{k}+8k=10+2\left(4k+\frac{1}{k}\right)$
$=10+2\left[{\left(2\sqrt{k}-\frac{1}{\sqrt{k}}\right)}^2+4\right]$
$\therefore$ Minimum value of $OP+OQ$ when
$2\sqrt{k}-\frac{1}{\sqrt{k}}=0\Rightarrow 2k-1=0\Rightarrow k=\frac{1}{2}$
$\therefore$ Minimum value $=10+8=18$
$\therefore y_{\min }=|a+b|$
$\text{ equation }\frac{x}{a}+\frac{y}{b}=1a>0,b>0$

$\frac{8}{a}+\frac{2}{b}=1$
$\frac{2}{b}=\left(1-\frac{8}{a}\right)=\frac{a-8}{a}$
$b=\frac{2a}{a-8}$
$z=a+\frac{2a}{a-8}=\left|\frac{a^2-8a+2a}{a-8}\right|=\left|\frac{a^2-6a}{a-8}\right|$
$(a-8)(4a+6)-\left(a^2-6a\right)=0$
$2a^2-44a+48-a^2+6a$
$a=|2|\text{ or }a=y$
$a^2-16a+48$
$(a-12)(ax-y)=0$